\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0,2\right)=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+0,2\right)=-\dfrac{891}{25}:4=-\dfrac{891}{100}\)

`(9/25 - 2.18 ) : ( 3 4/5 + 0,2)`

`=(9/25 - 36):( 19/5 + 1/5)`

`=(9/25 - 900/25):4`

`=-891/25 : 4`

`=-891/25 . 1/4`

`=-891/100`

Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3+\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):4\)

\(=\dfrac{-891}{25}:4\)

\(=\dfrac{-891}{100}\)

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

\(=4-1-\dfrac{2}{5}\)

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0,2\right)\)

=\(\left(\dfrac{9}{25}-36\right):4\)

= -35,64:4=-8,91

Mình chỉ cho đáp án thôi,sai thì châm chước cho mìn nha!

a)-91 phần200

b)-25phần 4

c)5 phần 2

d)2

e)0

a, ( 0,36-2,18) : ( 3,8 + 0,2)

= -1,82 : 4

=-0,455 hay -91/200

b, 3/8*19/1/3-3/8*33/1/3

=3/8*(19/1/3-33/1/3)

=3/8*(-14)

=-21/4

sorry

Giải:

a) \(\left(\dfrac{9}{25}-2.18\right):\left(\dfrac{19}{5}+0,2\right)\)

\(=\left(-\dfrac{891}{25}\right):4\)

\(=-\dfrac{891}{100}\)

Vậy ...

b) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=1+\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1+0,5\right)+\left(\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)\)

\(=1,5+0+1\)

\(=2,5\)

Vậy ...

a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)

b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)

c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)

d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)

a)

b)

a)

\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(-1.82\right):4=-0.455\)

b) \(\dfrac{5}{18}-1.456:\dfrac{7}{25}+4.5\cdot\dfrac{4}{5}\)

\(=\dfrac{5}{18}-5.2+3.6=-1\dfrac{29}{90}\)